QA #2 Series

Questions and Answers Sheet 2 Series Question #1 A pair of dice is rolled until a sum of either 5 or 7 appears. Find the probability that a 5 occurs first. Hint: Let 𝐸𝑛 denote the event that a 5 occurs on the nth roll and no 5 or 7 occurs on the first n-1 rolls. Compute 𝑃(𝐸𝑛 ) and argue that ∑ 𝑛 = 1∞ 𝑃(𝐸𝑛 ) is the desired probability. Answer: Experiment: Throwing two fair six sided dice. Use the notation from the exercise: 𝐸𝑛 - Event that a sum of 5 is rolled in the n-th roll, and neither 5 or 7 were rolled before. Now every event where the sum of 5 is rolled before 7 is some 𝐸𝑛 - because the checking if this event happened in the experiment is: from the begging watch for sum of 5 or 7, if the sum of 5 appeared first both sums, lets say in n-th roll this is event 𝐸𝑛 We have 36 equally likely possibilities for every dice roll, this means a total of 36𝑛 possibilities. 4 of these 36 − (1,4), (4,1), (2,3), (3,2) are possible if the sum is 5 and 6 are possible if the sum is 7. This leaves 26 possible rolls that sum up to a number that's not 5 or 7. If 𝐸𝑛 happened, the first n-1 rolls are some from 26𝑛−1 vectors and the last one is one of 4 rolls. This ciunting uses the basic principle of counting that later translates to indepence. 𝑃(𝐸𝑛 ) = 26𝑛−1 ⋅4 36𝑛 4 26 𝑛−1 = 36 (36) 1 13 𝑛−1 1 13 𝑛−1 = 9 (18) ∞ ∑∞ 𝑛=1 𝑃(𝐸𝑛 ) = ∑𝑛=1 9 (18) 13 𝑘 1 = 9 ∑∞ 𝑘=0 (18) 1 1 1 1− 18 1 =9⋅ =9⋅ 1 =9⋅ = 2 13 5 18 18 5 5 Result: 2 5 Question #2 Prove that 1 · 1! + 2 · 2! +··· +𝑛 · 𝑛! = (𝑛 + 1)! − 1 whenever n is a positive integer. Answer: To proof 1 ⋅ 1! + 2 ⋅ 2! + ⋯ + 𝑛 ⋅ 𝑛! = (𝑛 + 1)! − 1 for every positive integer n. Proof by induction Let 𝑃 (𝑛) be 1 ⋅ 1! + 2 ⋅ 2! + ⋯ + 𝑛 ⋅ 𝑛! = 1 ⋅ 1 = (𝑛 + 1)! − 1 Basis step n=1 1 ⋅ 1! + 2 ⋅ 2! + ⋯ + 𝑛 ⋅ 𝑛! = 1 ⋅ 1! = 1 ⋅ 1 = 1 (𝑛 + 1)! − 1 = (1 + 1)! − 1 = 2! − 1 = 2 − 1 = 1 We then note P(1) is true. Induction step let 𝑃(𝑘 + 1) is also true 1 ⋅ 1! + 2 ⋅ 2 + ⋯ + 𝑘 ⋅ 𝑘! = (𝑘 + 1)! − 1 We need to prove that 𝑃(𝑘 + 1) is also true. 1 ⋅ 1! + 2 ⋅ 2! + ⋯ + 𝑘 ⋅ 𝑘! + (𝑘 + 1) ⋅ (𝑘 + 1)! = ( 𝑘 + 1 )! − 1 + (𝑘 + 1 ) ⋅ (𝑘 + 1 )! = 1 ⋅ (𝑘 ⋅ 1)! − 1 + (𝑘 + 1 ) ⋅ (𝑘 + 1)! = 1 ⋅ (𝑘 + 1)! + (𝑘 + 1) ⋅ ( 𝑘 + 1)! − 1 = (1 + 𝑘 + 1)(𝑘 + 1)! − 1 = (𝑘 + 2)! − 1 = ((𝑘 + 1) + 1)! − 1 We then note that 𝑃 (𝑘 + 1) is also true Question #3 Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do not show that 𝑅𝑛(𝑥 ) → 0 .] Also find the associated radius of convergence. 𝑓(𝑥) = 2𝑥 Answer: Find a few derivatives, and calculate their values at a=0. 𝑓 (𝑥 ) = 2 𝑥 𝑓 (0) = 1 𝑓(𝑥 ) = (ln 2)2𝑥 𝑓 ′ (0) = ln 2 𝑓(𝑥 ) = (ln 2)2 2𝑥 𝑓 ′′ (0) = (ln 2)2 𝑓(𝑥 ) = (ln 2)3 2𝑥 𝑓 ′′′ (0) = (ln 2)3 𝑓 (4) (𝑥 ) = (ln 2)4 2𝑥 𝑓 (4) (0) = (ln 2)4 Plug everything into the Maclaurin general form 𝑓 (𝑥 ) = 𝑓 (0) + 𝑓 (𝑥 ) = 1 + ln 2 1! 𝑓′ (0) 1! 𝑥+ 𝑥+ 𝑓′′ (0) (ln 2)2 2! 2! 𝑥2 + 𝑥2 + 𝑓′′′ (0) (ln 2)3 3! 3! 𝑥3 + ⋯ 𝑥3 + (ln 2)4 4! 𝑥4 + ⋯ Find the pattern of the numbers to erite in summation form 𝑓(𝑥 ) = ∑∞ 𝑛=0 (ln 2)𝑛𝑥 𝑛 𝑛! Use the Ratio Test 𝑎𝑛 = (ln 2)𝑛𝑥 𝑛 𝑛! an +1 (ln 2)n +1 x n +1 n! (ln 2) x | |=|  |=| | n n an (n + 1)! (ln 2) x n +1 lim | n → (ln 2) x |= 0 n +1 The limit is <1 for all x so it converges for all real numbers and the radius of converges is lim( k +0) k ) / (6k 4 + 1) = 0 𝑅=∞ Question #4 Express the definite integral as an infinite series and find its value to within an error of at most 10−4 1 ∫0 cos(𝑥 2 ) 𝑑𝑥 Answer: Term-by-Term Differentiation and Integration 𝑛 𝐹(𝑥 ) = ∑∞ 𝑛=0 𝑎𝑛 (𝑥 − 𝑐 ) has radius of convergence 𝑅 > 0 . Then F is differentiable on (𝑐 − 𝑅, 𝑐 + 𝑅). Furtehermore, we can integrate and differentiate term by term. For x\in(c-R,c+R) 𝑛−1 𝐹(𝑥 ) = ∑∞ 𝑛=1 𝑛𝑎𝑛 (𝑥 − 𝑐 ) 𝑎 𝑛 𝑛+1 (A any constant) ∫ 𝐹(𝑥 )𝑑𝑥 = 𝐴 + ∑∞ 𝑛=0 𝑛+1 (𝑥 − 𝑐 ) These series have the same radius of convergence R 1 Here we need to find the value of 𝐹 (𝑥 ) = ∫0 cos(𝑥 2 ) 𝑑𝑥 we will use the expansion of the cos 𝑥 𝑥 2𝑛 𝑛 From table 2, We have Maclaurin Series 𝑓 (𝑥 ) = cos 𝑥 = ∑∞ 𝑛=0(−1) (2𝑛)! = 1 − 𝑥2 2! + 𝑥4 4! − converges for for all x here x is replaced with 𝑥 2 𝑛 cos 𝑥 2 = ∑∞ 𝑛=0(−1) (𝑥 2 ) 2𝑛 (2𝑛)! 𝑥 4𝑛 𝑛 cos 𝑥 2 = ∑∞ 𝑛=0(−1) (2𝑛)! 𝑑𝑥 1 1 \ ∫0 cos 𝑥 2 𝑑𝑥 = ∑𝑛=0 𝑖𝑛𝑓𝑡(−1)𝑛 (4𝑛+1)(2𝑛)! Now we need to find out F(1) error less than 0.0001 1 1 𝑛 ∫0 cos 𝑥 2 𝑑𝑥 = ∑∞ 𝑛=0(−1) (4𝑛+1)(2𝑛)! 1 Above is alternating series with 𝑎𝑛 = (4𝑛+1)(2𝑛)! 1 Using the error the bounded of alternating series |∫0 cos 𝑥 2 𝑑𝑥 − 𝑆𝑁 | < 𝑎𝑁+1 1 𝑎𝑁+1 = (4𝑁+5)(2𝑁+2)! < 0.0001 ⇒ (4𝑁 + 5)(2𝑁 + 2)! > 10000 for N=2 we have (4𝑁 + 5)(2𝑁 + 2)! = 13 × 6! = 9360 < 10000 for N=3 we have (4𝑁 + 5)(2𝑁 + 2)! = 17 × 8! = 685440 > 10000 so we choose N=3 1 ∫0 cos 𝑥 2 𝑑𝑥 ≈ ∑𝑛 1 ∫ cos 𝑥 2 𝑑𝑥 = 1 − 0 1 = 0∞ (−1)𝑛 (4𝑛+1)(2𝑛)! 1 1 1 + − 5 × 2! 9 × 4! 13 × 6! 1 ∫0 cos 𝑥 2 𝑑𝑥 = 0.904522792 1 Result: ∫0 cos 𝑥 2 𝑑𝑥 = 0.904522792 Question #5 Approximate the sum of the series correct to four decimal places. 𝑥6 6! +⋯ ∑∞ 𝑛 𝑛 = 1(−1)𝑛 3𝑛𝑛! Answer: We know that, in an alternating series, the estimate error is smaller than the first neglected term. That's why we will apply. We can see that we have: 1 |𝑅𝑛 | ≤ 𝑏𝑛+1 = 𝑛+1 < 0.00001 (𝑛+1)! 3 Since this can be a tough inequality to solve, we can use graphing calculator to figure out for which n this will aply. We can see that we have: 1 35 5! 1 ≈ 0.00003 > 0.00001 > 36 6! ≈ 0.000002 Therefore, we need to add at least 5 terms for the error to be less than 0.00001 , that is, for our approximation to be correct to the fourth decimal. Now we have: 𝑠5 = ∑5𝑘=1 (−1)𝑘 3𝑘 𝑘! ≈ −0.28347 The sum of the series is approximately −0.28347 Question #6 find a formula for an for the arithmetic sequence. 𝑎1 = −4, 𝑎5 = 16 Answer: Generic formula for an arithmetic sequence in which nth term equals the first term plus the common difference times the number of differences between the first term and the nth term: 𝑎𝑛 = 𝑎1 + 𝑑(𝑛 − 1) , then 1) 𝑎5 = 𝑎1 + 4𝑑 16 = −4 + 4𝑑 𝑑 = 5 2) 𝑎𝑛 = −4 + 5(𝑛– 1) 𝑎𝑛 = −4 + 5𝑛 – 5 = 5𝑛 − 9 Question #7 Write the first five terms of the arithmetic sequence. 𝑎1 = 5, 𝑑 = 6 Answer: To find 5 terms, add 6 to each term. 𝑎2 = 𝑎1 + 𝑑 = 5 + 6 = 11, 𝑎1 = 5; 𝑑 = 6 𝑎3 = 𝑎2 + 𝑑 = 11 + 6 = 17, 𝑎2 = 11; 𝑑 = 6 𝑎4 = 𝑎3 + 𝑑 = 17 + 6 = 23, 𝑎3 = 17; 𝑑 = 6 𝑎5 = 𝑎4 + 𝑑 = 23 + 6 = 29, 𝑎4 = 23; 𝑑 = 6. The answer is: 5, 11, 17, 23, 29. Question #8 Which of the following series are geometric? Which are power ones? a) 1 + 𝑥/2 + 𝑥 2 /4 + 𝑥 3 /8 + 𝑥 4 /16 + ⋯ b) 1 + 1.1 + 1.21 + 1.331 + 1.4641 + 1.6105 + … c) (1/3)2 + (1/3)4 + (1/3)6 + (1/3)8 + ⋯ d) 1 + 𝑥 + 𝑥 2 /(2!) + 𝑥 3 /(3!) + 𝑥 4 /(4!) + ⋯ e) 1 + 1/2 + 1/3 + 1/4 + 1/5 + … f) 1/𝑥 2 + 1/𝑥 + 1 + 𝑥 + 𝑥 2 + 𝑥 3 + 𝑥 4 + ⋯ Answer: a) 1 + 𝑥/2 + 𝑥 2 /4 + 𝑥 3 /8 + 𝑥 4 /16 + ⋯ is both power and geometric series. b) 1 + 1.1 + 1.21 + 1.331 + 1.4641 + 1.6105 + … is only power series c) (1/3)2 + (1/3)4 + (1/3)6 + (1/3)8 + ⋯ is geometric series. d) 1 + 𝑥 + 𝑥 2 /(2!) + 𝑥 3 /(3!) + 𝑥 4 /(4!) + ⋯ is geometric series. e) 1 + 1/2 + 1/3 + 1/4 + 1/5 + … is neither geometric nor power one. f) 1/𝑥 2 + 1/𝑥 + 1 + 𝑥 + 𝑥 2 + 𝑥 3 + 𝑥 4 + ⋯ is power series. Question #9 The following series ∑𝑜(𝑘=0) 𝑜1/𝑘 5 is: a) alternating series b) convergent p-series c) divergent p-series d) geometric series Answer: ∑𝑜(𝑘=0) 𝑜1/𝑘 5 This series is convergent p-series. Since ∑𝑜(𝑘=0) 𝑜1/𝑘 5 converges for 𝑝 > 1 and diverges for 𝑝 ≤ 1 So compare this then: 𝑝 = 5 > 1 Thus, given series converges by p-series test. Question #10 Consider two series ∑𝑜(𝑛=1) 𝑜1/2𝑝𝑖𝑛 and ∑𝑜(𝑛=0) 𝑜1/𝑝𝑖 2 𝑛. Which of them converges? Answer: Series A: ∑𝑜(𝑛=1) 𝑜1/2𝑝𝑖𝑛 = 1/2𝑝𝑖1/1 + 1/2 + 1/3 + Harmonic series – Divergent Series B:∑𝑜(𝑛=0) 𝑜1/𝑝𝑖 2 𝑛 = (1/𝑝𝑖 2 ) + (1/𝑝𝑖 2 )2 + Geometric series, so 1/𝑝𝑖 2 < 1 – Convergent Thus, only series B converges. Question #11 Name the series: a) ∑(𝑛 = 0)𝑜 𝑜(𝑓 𝑛 𝑥𝑥𝑎)/(𝑛!)𝑥𝑥(𝑥– 𝑎)𝑛 b) ∑(𝑛 = 0)𝑜 𝑜(𝑓 𝑛 𝑥𝑥0)/(𝑛!)𝑥𝑥𝑥 𝑛 Answer: a) ∑(𝑛 = 0)𝑜 𝑜(𝑓 𝑛 𝑥𝑥𝑎)/(𝑛!)𝑥𝑥(𝑥– 𝑎)𝑛 is a Taylor series. b) ∑(𝑛 = 0)𝑜 𝑜(𝑓 𝑛 𝑥𝑥0)/(𝑛!)𝑥𝑥𝑥 𝑛 is a Maclaurin series. Question #12 Classify the following series: 1 + 1/4 + 1/9 + 1/16 + 1/25 + … Answer: 1 + 1/4 + 1/9 + 1/16 + 1/25 + … 1/12 + 1/22 + 1/32 + 1/42 + 1/52 + ⋯ , so ∑𝑜(𝑛=1) 𝑜1/(𝑛)2 Harmonic series are given as: ∑𝑜(𝑛=1) 𝑜1/(𝑛)𝑝 , in the above series p = 2 Hence, the series would be classified as a p-series with 𝑝 > 1 Question #13 Classify whether this series converges ∑(𝑘 = 1)𝑜 𝑜((−1)𝑘 𝑥𝑥𝑘)/(6𝑘4 + 1) Answer: ∑(𝑘 = 1)𝑜 𝑜((−1)𝑘 𝑥𝑥𝑘)/(6𝑘4 + 1) ak = (k / (6k 4 + 1) bk = (k / (6k 4 + 1) 1) 𝑏𝑘 ≥ 0 2) lim( k +0) k ) / (6k 4 + 1) = 0 Then, by alternative series test we can converge. The term of series are alternative and limit of absolute value is 0, so series converge by alternative series test. Question #14 Find the sum of the series, if it converges: a) ∑(𝑘 = 0)𝑜 𝑜(1/3)𝑘 b) ∑(𝑘 = 0)𝑜 𝑜(−1/2)𝑘 Answer: a) ∑(𝑘 = 0)𝑜 𝑜(1/3)𝑘 A series of form 𝑟 𝑘 converges for |𝑟| < 1 Here, 𝑟 = 1/3 < 1 , so the series converges. ∑(𝑘 = 0)𝑜 𝑜(1/3)𝑘 = 1/(1– (1/3)) = 1/(2/3) = 3/2 b) ∑(𝑘 = 0)𝑜 𝑜(−1/2)𝑘 Here, 𝑟 = −1/2 , so the series converges. ∑(𝑘 = 0)𝑜 𝑜(−1/2)𝑘 = 1/(1– (−1/2)) = 1/(1 + 1/2) = 2/3 Question #15 Determine whether the series converges: ∑(𝑘 = 1)𝑜 𝑜4/(𝑘 + 2)5 Answer: ∑(𝑘 = 1)𝑜 𝑜4/(𝑘 + 2)5 𝑎𝑘 = 4/(𝑘 + 2)5 a_k = 4/(k^5(1 + 2/k)^5) Let assume that, 𝑏𝑘 = 1/𝑘5 By the limit comparison test, 𝐿 = 𝑙𝑖𝑚(𝑘→𝑜𝑜)𝑎𝑘 /𝑏𝑘 𝐿 = 𝑙𝑖𝑚(𝑘→𝑜𝑜)(4/(𝑘5 (1 + 2/𝑘)5 ))/(1/𝑘 5 ) 𝐿 = 𝑙𝑖𝑚(𝑘→𝑜𝑜)4/(1 + 2/𝑘)5 𝐿 = 4(1 + 2/𝑜𝑜)5 𝐿 = 4/(1 + 0)5 𝐿 = 4 Here 𝐿 = 4 – finite and non-zero, therefore the series is a p-series with 𝑝 = 5(𝑏𝑘 = 1/𝑘5 ) , so the given series is converges by the properties of p-series. Question #16 Determine whether the following series converges: ∑(𝑘 = 1)𝑜 𝑜(2/(𝑘 + 5)3 ) . Answer: ∑(𝑘 = 1)𝑜 𝑜(2/(𝑘 + 5)3 ) lim (𝑎𝑘 ) = (𝑘→𝑜𝑜) lim (2/(𝑘 + 5)3 ) = 9/𝑜𝑜 = 0 ((𝑘→5)5 ) The limit of the terms of the series is 0, so the series diverges by the divergent test. Question #17 Determine whether the following series converges if “a” is real: ∑(𝑘 = 1)𝑜 𝑜(1– 𝑎/𝑘)6 𝑘 Answer: ∑(𝑘 = 1)𝑜 𝑜(1– 𝑎/𝑘)6 𝑘 𝑎𝑛 ! = 0, then ∑(𝑘 = 1)𝑜 𝑜𝑎𝑘 diverges. lim if (𝑛→𝑜𝑜) Here 𝑎𝑘 = (1– 𝑎/𝑘)(𝑘/6) So lim (𝑘→𝑜𝑜) 𝑎𝑘 = = lim (1– 𝑎/𝑘)(𝑘/6) = (𝑘→𝑜𝑜) = = lim (𝑘/6) 𝑥𝑥(−𝑎/𝑘) = (𝑒 (𝑘→𝑜𝑜) ) lim (𝑒 (𝑘→𝑜𝑜) ) lim (𝑘→𝑜𝑜) − 𝑎/6 = 𝑎𝑘 = 𝑒 (−𝑎/6) a is real number, so 𝑒 (−𝑎/6) ! = 0 for real number a. lim (𝑘→𝑜𝑜) 𝑎𝑘 ! = 0 Thus, by divergent test, given series is divergent. Question #18 Determine whether the following series converges or diverges: ∑(𝑛 = 1)𝑜 𝑜(𝑛3 + 𝑛4 )/𝑛5. Answer: ∑(𝑛 = 1)𝑜 𝑜(𝑛3 + 𝑛4 )/𝑛5 = = ∑(𝑛 = 1)𝑜 𝑜1/𝑛2 + ∑𝑜(𝑛=1) 𝑜1/𝑛 As ∑(𝑛 = 1)𝑜 𝑜1/𝑛2 is convergent by p-series test and ∑𝑜(𝑛=1) 𝑜1/𝑛 is divergent, Series diverges because we can break the series up into a convergent p-series and a divergent p-series. Question #19 Determine whether the following series converges or diverges: ∑(𝑘 = 1)𝑜 𝑜3/(𝑘 + 4)3 Answer: ∑(𝑘 = 1)𝑜 𝑜3/(𝑘 + 4)3 = = ∑(𝑘 = 1)𝑜 𝑜3/𝑘3 By p-series test ∑(𝑘 = 1)𝑜 𝑜1/ℎ𝑝 , 𝑝 > 1 – converges, 𝑝 < 1 – diverges. So this series converges, cause it’s a p-series with p = 3. Question #20 Please, describe the application of the Alternating Series Test for each series: a) ∑(𝑛 = 1)𝑜 𝑜(−1)𝑛 /𝑛 b) ∑(𝑛 = 1)𝑜 𝑜1/𝑛 c) ∑(𝑛 = 1)𝑜 𝑜(−1)𝑛 Answer: a) ∑(𝑛 = 1)𝑜 𝑜(−1)𝑛 /𝑛 – the given series converges b) ∑(𝑛 = 1)𝑜 𝑜1/𝑛 – the Alternatin Series Test is inconclusive or cannot be applied to the series c) ∑(𝑛 = 1)𝑜 𝑜(−1)𝑛 – the Alternatin Series Test is inconclusive or cannot be applied to the series